Rational even polynomials maximally tangent to the unit circle

Question

This question is motivated by College Mathematics Journal problem 1196, proposed by Ferenc Beleznay and Daniel Hwang. My solution to this problem (pre-publication version here) uses Chebyshev polynomials to show that for every $m \in \mathbb{N}$ there is an even polynomial $P_{2m}(x) \in \mathbb{Q}[x]$ of degree $2m$ such that the graph

$$\{(x, P_{2m}(x)) : -1 \le x \le 1\}$$

is tangent to the unit circle $x^2+y^2=1$ at $2m-1$ points, and has two further points of intersection at $\pm 1$. Numerical calculations suggest that something stronger is true.

Is is true that for every $m \in \mathbb{N}$ with $m \ge 2$ there is an even polynomial $Q_{2m}(x) \in \mathbb{Q}[x]$ of degree $2m$ such that the graph $\{(x, Q_{2m}(x)) : -1 \le x \le 1\}$ is tangent to the unit circle at $2m-3$ points, and has two further triple points of intersection?

Examples. The quartic $Q_2(x) = 1 - \frac{(3x)^2}{2} + \frac{(3x)^4}{24}$ is tangent to the unit circle at $(x,y) = (0,1)$ and has triple points at $(x,y) = (\pm \frac{2\sqrt{2}}{3},-\frac{1}{3})$. The graphs below show the intersections for

\begin{align*} P_8(x) &= 1 - 25 x^2 + 104 x^4 - 144 x^6 + 64 x^8 \\ Q_8(x) &= 1 - \frac{(7 x)^2}{2} + \frac{(7 x)^4}{24} - \frac{(7 x)^6}{864} + \frac{(7 x)^8}{96768} \end{align*}

Further remarks.

1. The polynomials $P_{2m}(x)$ in fact have coefficients in $\mathbb{Z}$. They meet the bound in Bezout's Theorem for the intersection multiplicity between the plane algebraic curves $y = P_{2m}(x)$ of degree $2m$ and $x^2 + y^2 = 1$ of degree $2$, since $(2m-1)\times 2 + 2 \times 1 = 4m$. If $Q_{2m}(x)$ exists then it also meets the bound, now with $(2m-3) \times 2 + 2 \times 3 = 4m$.
2. I've asked for a rational polynomial $Q_{2m}(x)$ because my numerical experiments suggest this is possible, and the coefficients of the polynomials I've found seem to have some interesting aritmetic properties, see the solutions below. (Does anyone recognise the denominators?) But I have not proved the existence of $Q_{2m}(x)$ working over the real numbers.

   \begin{align*} 2:\quad & 1 - \frac{(3x)^2}{2} + \frac{(3x)^4}{24} \\ 3:\quad & 1 - \frac{(5x)^2}{2} + \frac{(5x)^4}{24} - \frac{(5x)^6}{1080} \\ 4:\quad & 1 - \frac{(7x)^2}{2} + \frac{(7x)^4}{24} - \frac{(7x)^6}{864} + \frac{(7x)^8}{96768} \\ 5:\quad & 1 - \frac{(9x)^2}{2} + \frac{(9x)^4}{24} - \frac{(9x)^6}{800} + \frac{(9x)^8}{64000} - \frac{(9x)^{10}}{14400000} \\ 6:\quad & 1 - \frac{(11x)^2}{2} + \frac{(11x)^4}{24} - \frac{7(11x)^6}{5400} + \frac{(11x)^8}{54000} - \frac{(11x)^{10}}{8100000} + \frac{(11x)^{12}}{3207600000} \end{align*} and for $m=7$ and $m=8$ there are $$\begin{align*} 7: \quad & \scriptstyle 1 - \frac{(13x)^2}{2} + \frac{(13x)^4}{24} - \frac{(13x)^6}{756} + \frac{(13x)^8}{49392} - \frac{(13x)^{10}}{6223392} + \frac{(13x)^{12}}{1568294784} - \frac{(13x)^{14}}{999003777408}\\ 8: \quad & \scriptstyle 1 - \frac{(15x)^2}{2} + \frac{(15x)^4}{24} - \frac{3(15x)^6}{2240} + \frac{15(15x)^8}{702464} - \frac{11(15x)^{10}}{59006976} + \frac{(15x)^{12}}{1101463552} - \frac{(15x)^{14}}{431773712384} + \frac{(15x)^{16}}{414502763888640}.\end{align*}$$ The prime factorizations of the denominators of the coefficients of $x^{2m}$ for $2 \le m \le 8$ are

   $$2^3\cdot 3, \quad 2^3\cdot 3^3\cdot 5,\quad 2^9\cdot 3^3\cdot 7,\quad 2^9\cdot 3^2\cdot 5^5, \quad 2^7\cdot 3^6\cdot 5^5\cdot 11,\quad 2^7\cdot 3^6\cdot 7^7\cdot 13, \quad 2^{25}\cdot 3\cdot 5\cdot 7^7$$

3. At Fedor Petrov's suggestion I computed the coordinates of the triple points. For $m \le 8$ the triple point with positive $x$-coordinate is

   $$\bigl(\frac{\sqrt{(2m+1)^2-1}}{2m+1}, \frac{(-1)^{m-1}}{2m+1}\bigr).$$

   (Sign of $y$ coordinate corrected from $(-1)^m$.) There appears to be no simple closed form for the double points in my solutions.

4. The original problem also required odd polynomials $P_{2m-1}$ of degree $2m-1$ with analogous tangency properties. I conjecture that the analogous question for $Q_{2m-1}$ also has a positive answer.

Answer 1

Denote $\varepsilon=\frac1{2m-1}$ (I guess it should be minus in the denominator, not plus, please recheck).

We want to find a polynomial $h_{m-1}(t)$ of degree $m-1$ and a polynomial $p_m(t)$ of degree $m$ such that $$p_m^2(t)-t(t-1+\varepsilon^2)h_{m-1}^2(t)=1-t,\quad\quad\quad\,\,\,\,\,\,\,\,(1)$$ $$h_{m-1}(1-\varepsilon^2)=0,\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(2)$$ $$\text{all roots of}\, h_{m-1}\, \text{are distinct and belong to}\, (0,1)\quad (3).$$ Indeed, if $\alpha$ is a root of $h_{m-1}$, then $(1)$ yields that the points $(\pm \sqrt{\alpha},p_m(\alpha))$ belong to a unit circle and to a graph of $y=p_m(x^2)$, and the tangency is at least double for all $\alpha$ and triple for $\alpha=\pm \sqrt{1-\varepsilon^2}$. Also $(0,p_m(1))$ is a point of tangency.

Now we look at (1) as Pell type equation $$\left(P-\sqrt{t(t-1+\varepsilon^2)}Q\right)\left(P+\sqrt{t(t-1+\varepsilon^2)}Q\right)=1-t.$$ There is a basic solution with 1 at right hand side: $P_0^2-t(t-1+\varepsilon^2)Q_0^2=1$ for $P_0=\alpha t-1$, $Q_0=\alpha$ with $\alpha:=\frac2{1-\varepsilon^2}$. A solution with $1-t$ in RHS is $(\beta t-1)^2-t(t-1+\varepsilon^2)\beta^2=1-t$, where $\beta(1\pm \varepsilon)=1$ (both signs are ok). So, we choose $$ p_m+\sqrt{t(t-1+\varepsilon^2)}h_{m-1}=\left(\beta t-1+\beta \sqrt{t(t-1+\varepsilon^2)}\right)\left(\alpha t-1+\alpha\sqrt{t(t-1+\varepsilon^2)}\right)^{m-1}. $$ For checking (2), we expand the brackets in RHS and choose only those which contain exactly one $\sqrt{t(t-1+\varepsilon^2)}$. In other words, $1-\varepsilon^2$ should be a root of $$(\alpha t-1)\beta+(m-1)\alpha(\beta t-1)=0.$$ Well, this happens if $\beta=(1+\varepsilon)^{-1}$ (straightforward check).

It remains to check (3). For this, we note that when $t$ goes from 0 to $1-\varepsilon^2$, the point $\alpha t-1+\alpha\sqrt{t(t-1+\varepsilon^2)}$ goes along the half of the unit circle. Denote $t=\cos \tau$ for $0<\tau<\pi$. We need $m-2$ distinct points $\tau\in (0,\pi)$ for which the imaginary part of $$\frac{\alpha}{\beta}\left(p_m+\sqrt{t(t-1+\varepsilon^2)}h_{m-1}\right)=(\cos \tau+i\sin \tau)^{m-1}\left(\cos \tau+i\sin \tau-\right(\frac{\alpha}{\beta}-1\left)\right)$$ equals to 0. As $\frac{\alpha}{\beta}-1=\frac{m}{m-1}$, this reads as $\sin m\tau=\frac{m}{m-1}\sin (m-1)\tau$, or $$\theta(\tau):=\frac{\sin (m-1)\tau}{\sin m\tau}=\frac{m-1}m.$$ For each $k=1,2,\ldots,m-2$ consider an interval $(\frac{\pi k}m,\frac{\pi(k+1)}m)$. In this interval the function $\theta(\tau)$ is continuous, and the limit values at the endpoints are $+\infty$ and $-\infty$ in some order. Thus $\theta$ has at least one root in each of these intervals, totally $m-2$ roots on $(0,\pi)$, as we need.